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2 years ago

Define Interquartile Range. I will check for plagiarism.

Mathematics

2 answers:

Setler [38]2 years ago

8 0

It’s the mean of the 50 percent of data in the middle in a given set of data. (Also called iqm)

nlexa [21]2 years ago

5 0

Interquartile range is the range is the number in the dead center, you have to divide the number line into 2 sections. The middle of everything and the middle of both section is the interquartile range

Hope this helps!

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Dishwasher detergent is sold in individual packets. The 20 packet box sells for $5.49. The 60 packet box sells for $10.97. The 9

poizon [28]

Answer:

60 packets

Step-by-step explanation:

Given that :

Unit cost of 20 packets

$5.49 / 20 =$0.2745

Unit cost of 60 packets :

$10.97 / 60 = $0.1828333

Unit cost of 90 packets :

$18.95 / 90 = $0.2105555

The 60 packets has the least cost per unit, hence it is the best buy.

5 0

3 years ago

Will give brainliest need anwser now: Karen and Kevin collect coins. Karen has x coins. Kevin has 4 coins fewer than four times

vodka [1.7K]

Answer:

100889999999999999999

5 0

3 years ago

48 cookies; 50% increase

saul85 [17]

Answer:

Step-by-step explanation:

48÷2= 24

48+24= 72

Hope this helps! :)

4 0

3 years ago

Read 2 more answers

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago

Use the given transformation to evaluate the integral. (10x + 10y) da r , where r is the parallelogram with vertices (−3, 12), (

Sever21 [200]

.......,,.............".............

5 0

3 years ago